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Let $Y_1$ and $Y_2$ be closed subspaces of a normed space $X$ (which may be infinite dimensional). How can I show that $Y_1$ and $Y_2$ must have different annihilators or else be the same subspace?

The annihilator $Y^0$ of a subspace $Y \subset X$ is defined as the set of functionals in $X'$ (the dual space of $X$) such that $f(y)=0\forall y \in Y, f \in Y^0$.

(Note: I have looked for similar questions, but all I have found are proofs that assume the space to be finite dimensional. I found a mention that this can be proved as a corollary to the Hahn-Banach theorem - how?)

Jessica Hansen
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1 Answers1

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If $Y_1 \neq Y_2$, there is some $y_1 \in Y_1$ with $y_1 \not\in Y_2$. By Hanh-Banach Theorem, there exists $f \in X^*$ such that $f(y_1) \neq 0$ and $f|_{Y_2}$ = 0. So $f \in (Y_2)^0$ but $f \not\in (Y_1)^0$.

user1101010
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  • Oh that is really nice. I was trying to think of a way to prove it, but I should have used hahn banach more carefully. – Andres Mejia Oct 26 '17 at 16:11
  • Hahn-Banach allows you to make an extension, but I don't understand how you automatically know that an extension exists such that $f(y_1) \neq 0$. – Jessica Hansen Oct 26 '17 at 16:43
  • @JohnnyHansen: This is a standard corollary of Hanh-Banach Theorem. In most textbooks, it is treated right after the exposition of Hanh-Banach Theorem. Check Rudin's Real and Complex Analysis or Folland's Real Analysis. – user1101010 Oct 26 '17 at 16:52